//递归算法

#include<iostream>
using namespace std;
int a[1000];
int oil(int n, int m)
{
	if (n == 0 || m < 0)
	    return 0;
	else
	{
		if (m == a[n])
			return 1;
		else
		{
			if (oil(n - 1, m - a[n]) == 1)
				return 1;
			if (oil(n - 1, m) == 1)
				return 1;
		}
	}
	return 0;
}
int main()
{
	int n, m;
	while (cin >> n >> m)
	{
		int tatol = 0;
		for (int i = 1; i <= n; i++)
		{
			cin >> a[i];
			tatol += a[i];
		}
		if (tatol < m || m==0)
			printf("no\n");
		else
		{
			if (oil(n, m))
				printf("yes\n");
			else
				printf("no\n");
		}
	}
	return 0;
}

//01背包

#include<algorithm>
#include<iostream>
using namespace std;
int dp[1000];

int main()
{
	int n, m,a;
	while (cin >> n >> m)
	{
		if (m == 0)
		{
			printf("no\n");
			continue;
		}
		memset(dp, 0, sizeof(dp));
		for (int i = 0; i < n; i++)
		{
			cin >> a;
			for (int j = m; j >= a; j--)
			dp[j] = max(dp[j], dp[j - a]+a);
		}
		printf("%s\n", dp[m] == m ? "yes" : "no");
	}
}

//bool背包

利用01背包的思想

只要j-a为1则说明可以放，利用位运算 | 得到answer
#include<iostream>
using namespace std;
bool dp[1000];
int main()
{
	int n, m,a;
	while (cin >> n >> m)
	{
		if (m == 0)
		{
			printf("no\n");
			continue;
		}
		memset(dp, 0, sizeof(dp));
		dp[0] = 1;

		for (int i = 0; i < n; i++)
		{
			cin >> a;
			for (int j = m; j >= a; j--)
				dp[j] = dp[j] | dp[j - a];
		}
		printf("%s\n", dp[m] ? "yes" : "no");
	}
	return 0;
}

//位运算

#include<iostream>
using namespace std;
int n, m;
int a[1000];
int main()
{
	while (cin >> n >> m)
	{
		int tatol = 0;
		for (int i = 1; i <= n; i++)
		{
			cin >> a[i];
			tatol += a[i];
		}
		if (tatol < m || m==0)
			printf("no\n");
		else
		{
			int t, t2 ;
			int lim = 1 << n;
			for (int i = 0; i < lim; i++)
			{
				t = 0, t2 = i;
				for (int j = 1; i <= n; j++)
				{
					if (t2 & 1)
						t += a[j];
					t2 >>= 1;
				}
				if (t == m)
				{
					printf("yes\n");
					return 0;
				}
			}
			if (t != m)
			{
				printf("no\n");
			}
		}
	}
	return 0;
}